Toán Lớp 9: P=(√x/√x-1–1/x-√x): (1/√x+1+2/x-1) với x>0; x≠1a) rút gọn P
b) Tính giá trị của P khi x=1/4
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Toán Lớp 9: P=(√x/√x-1–1/x-√x): (1/√x+1+2/x-1) với x>0; x≠1a) rút gọn P
b) Tính giá trị của P khi x=1/4
Comments ( 1 )
a)P = \dfrac{{x – 1}}{{\sqrt x }}\\
b) – \dfrac{3}{2}
\end{array}\)
a)P = \left( {\dfrac{{\sqrt x }}{{\sqrt x – 1}} – \dfrac{1}{{x – \sqrt x }}} \right):\left( {\dfrac{1}{{\sqrt x + 1}} + \dfrac{2}{{x – 1}}} \right)\\
= \dfrac{{x – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}:\left( {\dfrac{{\sqrt x – 1 + 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}} \right)\\
= \dfrac{{x – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}\\
= \dfrac{{x – 1}}{{\sqrt x }}\\
b)Thay:x = \dfrac{1}{4}\\
\to P = \dfrac{{\dfrac{1}{4} – 1}}{{\sqrt {\dfrac{1}{4}} }} = – \dfrac{3}{2}
\end{array}\)